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Sum_{0<j<n} n^3-j^3.
3

%I #20 Jun 17 2017 03:06:25

%S 7,45,156,400,855,1617,2800,4536,6975,10285,14652,20280,27391,36225,

%T 47040,60112,75735,94221,115900,141120,170247,203665,241776,285000,

%U 333775,388557,449820,518056,593775,677505,769792,871200,982311

%N Sum_{0<j<n} n^3-j^3.

%C a(n) = n^4-p(n), where p(n) is the n-th partial sum of (j^3).

%C a(n) = t(n)-t(n-1), where t = A206809.

%C For a guide to related sequences, see A206817.

%H Danny Rorabaugh, <a href="/A206808/b206808.txt">Table of n, a(n) for n = 2..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n) = (3*n^4-2*n^3-n^2)/4. G.f.: -x^2*(x^2+10*x+7) / (x-1)^5. - _Colin Barker_, Jul 11 2014

%e a(2) = 2^3-1^3 = 7.

%e a(3) = (3^3-1^3) + (3^3-2^3) = 45.

%t s[k_] := k^3; t[1] = 0;

%t p[n_] := Sum[s[k], {k, 1, n}];

%t c[n_] := n*s[n] - p[n];

%t t[n_] := t[n - 1] + (n - 1) s[n] - p[n - 1]

%t Table[c[n], {n, 2, 50}] (* A206808 *)

%t Flatten[Table[t[n], {n, 2, 35}]] (* A206809 *)

%o (PARI) vector(100, n, (3*n^4+10*n^3+11*n^2+4*n)/4) \\ _Colin Barker_, Jul 11 2014

%o (PARI) Vec(-x^2*(x^2+10*x+7)/(x-1)^5 + O(x^100)) \\ _Colin Barker_, Jul 11 2014

%o (Sage) [sum([n^3-j^3 for j in range(1,n)]) for n in range(2,35)] # _Danny Rorabaugh_, Apr 18 2015

%Y Cf. A206817, A206806.

%K nonn,easy

%O 2,1

%A _Clark Kimberling_, Feb 15 2012